Localisation and completion of $\mathbb{Z}$ This question might seem a little vague, but any information will be helpful.
Fix a prime $p$ in $\mathbb{Z}$. It is easy to see as rings that the localisation at $(p)$ is contained in the $(p)$-adic completion (i.e. $\mathbb{Z}_{(p)}\subset\mathbb{Z}_p$), but what other elements can we claim is in the completion? Can we say something like $\sqrt[n]{q} \in\mathbb{Z}_p$ for any prime $q$ other than $p$ (probably by using Hensel's lemma)?
Edit: As suggested in the comments, this older question gives a good insight into the answer.
The question came up because someone told me that the maximal ideal $m_{O_\mathbb{C_p}}$ (the maximal ideal of the ring of integers of $\mathbb{C}_p$) is generated by $\{\sqrt[n]{p}: n\in\mathbb{N}\}$. Does that mean that $\mathbb{Q}_p[\{\sqrt[n]{p}: n\in\mathbb{N}\}]=\mathbb{Q}_p^{alg}$ (where $\mathbb{Q}_p^{alg}$ is the algebraic closure of $\mathbb{Q}_p$)?
Again, I know the question is pretty vague, but thanks in advance for any help!
 A: $\Bbb{Z}_p$ is uncountable, it contains a lot of elements non-algebraic over $\Bbb{Q}$.
For the elements algebraic over $\Bbb{Q}$ they are a bit complicated to describe. What we can say

*

*A number field $L$ embeds into $\Bbb{Z}_p$ iff its ring of integers has an unramified prime ideal above $p$ and whose redidue field is $\Bbb{F}_p$.


*Also $\Bbb{Q}_p\cap \overline{\Bbb{Q}}$ is generated over $\Bbb{Q}$ by some elements for which Hensel lemma holds. More precisely $f\in \Bbb{Q}[x]$ irreducible has a root in $\Bbb{Z}_p$ iff
there is $g\in \Bbb{Z}[x]$ monic irreducible with a simple root in $\Bbb{Z}/(p)$ and such that $f$ has a root in $\Bbb{Q}[x]/(g)$.
For your last question, no.
$\mathbb{Q}_p^{alg} = K_\infty$ where $F=\Bbb{Q}_p[\zeta_{p^\infty-1}], K_0=F[p^{1/(p^\infty-1)}], K_{n+1}=  K_n[K_n^{1/p}]$.
$F$ is the maximal unramified extension, $K_0$ is the maximal tamely ramified extension, and the solvability of the ramification group plus Kummer theory gives that $K_\infty$ is algebraically closed.
The maximal ideal of $O_{\Bbb{C}_p}$ is generated by the $p^{1/n}$ just because it is the valuation ring of a $\Bbb{Q}$-valued valuation.
